When defining class method, why there an equals sign in argument e.g. func(foo=3)

Question

I'm learning python, trying to figure out how does this code actually work:

def func(**args):
    class BindArgs(object):
       foo = args['foo']
       print 'foo is ', foo
       def __init__(self,args):
           print "hello i am here"
    return BindArgs(args) #return an instance of the class

f = func(foo=2)

Output:

foo is 2 hello i am here

But it's very confused that in the argument of a function func(foo=2) that takes equation mark in it. Could you please explain how the flow works?

The answer from the link is that: in my example, `args == 'foo' ` and `args['foo'] == 2`. Quick question, if class `GindArgs(object)` could take type `**args` as argument, does that mean regular class could actually take "keyword (dictionary)" as argument too? — Jason, Dec 06 '17 at 22:33
Yes, functions and therefore class constructions can get a dictionary as their arguments. — Happy Ahmad, Dec 07 '17 at 10:17

score 1 · Answer 1 · answered Dec 06 '17 at 22:20

Here is an abstract: You call the function func and pass a dictionary as the argument to it. In the func function, you define a class named BindArgs, and then in the return statement, you first make an object (instance) from the BindArgs class and then return that instance. Notice that the statement foo = args['foo'] get the key value of 'foo' from the args dictionary (that is 2 in your sample code). the init also will be run as the constructor of the class when you are creating an object.

When defining class method, why there an equals sign in argument e.g. func(foo=3)

1 Answers1